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(C) Given equation: $4 v^2 = 25 - x^2$. Divide by $4$: $v^2 = \frac{25}{4} - \frac{x^2}{4}$. Comparing this with the standard equation of simple harmo

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$4 \pi$

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(C) Given equation: $4 v^2 = 25 - x^2$. Divide by $4$: $v^2 = \frac{25}{4} - \frac{x^2}{4}$. Comparing this with the standard equation of simple harmonic motion $v^2 = \omega^2 (A^2 - x^2)$,we get $v^2 = \frac{1}{4} (25 - x^2)$. Here,$\omega^2 = \frac{1}{4}$,which implies $\omega = \sqrt{\frac{1}{4}} = \frac{1}{2} 	ext{ rad/s}$. The time period $T$ is given by $T = \frac{2 \pi}{\omega}$. Substituting the value of $\omega$: $T = \frac{2 \pi}{1/2} = 4 \pi 	ext{ s}$.

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4 v^2 = 25 - x^2$,then the time period is

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